I have two lists:
list_1 = [1,1, 2,2,2, 3,3, 4, 4, 4 ,4, 4, 5,5,5,5]
list_2 = [5, 5, 5, 6, 6, 7]
I want to return the list of elements that appear only in the first list but not in the second and the list should be sorted ascending so the result is like this:
[1, 3, 2, 4]
So far I have this:
def sorted_nums(list_1,list2_2):
c = (set(list_1) - set(list_2))
d = dict.fromkeys(c, 0)
for index in list_1:
if index in c:
d[index] =1
return d
a = sorted_nums(list_1,list_2)
b = sorted(a.items(), key = lambda x: x[1])
print(b)
and it returns this:
[(1,2), (3,2), (2,3), (4,5)]
Could you help me to change the last part of the code so that I get the result I want?
CodePudding user response：
You just want to sort by the occurance (counts) of lis1, taking the set-difference from the second list:
>>> list1 = [1,1, 2,2,2, 3,3, 4, 4, 4 ,4, 4, 5,5,5,5]
>>> list2 = [5, 5, 5, 6, 6, 7]
>>> from collections import Counter
>>> counts1 = Counter(list1)
>>> sorted(counts1.keys() - set(list2), key=counts1.get)
[1, 3, 2, 4]
CodePudding user response：
use set operations
a = [1,1, 2,2,2, 3,3, 4, 4, 4 ,4, 4, 5,5,5,5]
b = [5, 5, 5, 6, 6, 7]
c = sorted(list(set(a) - set(b)))
print(c)
CodePudding user response：
Method 1:
By using sorted(array, key = list_1.count) to sort by occurrence but this will be slow.
list_1 = [1,1, 2,2,2, 3,3, 4, 4, 4 ,4, 4, 5,5,5,5]
list_2 = [5, 5, 5, 6, 6, 7]
def sorted_nums(list_1,list2_2):
return sorted(list(set(list_1) - set(list_2)), key = list_1.count)
sorted_nums(list_1, list_2)
Method 2 (Credits):
By using Counter. This is a faster approach. :
from collections import Counter
list_1 = [1,1, 2,2,2, 3,3, 4, 4, 4 ,4, 4, 5,5,5,5]
list_2 = [5, 5, 5, 6, 6, 7]
counter_1 = Counter(list1)
def sorted_nums(list_1,list2_2):
return sorted(counter_1.keys() - set(list2), key=counter_1.get)
sorted_nums(list_1, list_2)
Output:
[1, 3, 2, 4]
CodePudding user response：
All you need is a map
def sorted_nums(list_1,list2_2):
c = (set(list_1) - set(list_2))
d = dict.fromkeys(c, 0)
for index in list_1:
if index in c:
d[index] =1
return d
list_1 = [1,1, 2,2,2, 3,3, 4, 4, 4 ,4, 4, 5,5,5,5]
list_2 = [5, 5, 5, 6, 6, 7]
a = sorted_nums(list_1,list_2)
b = list(map(lambda x:x[0],sorted(a.items(), key = lambda x: x[1]))) ## changed here
print(b)
[1, 3, 2, 4]
Apply function lambda x:x[0] over each and every elements using map
CodePudding user response：
You can do it with Python built-in manipulation functions:
def sorted_nums(list_1,list_2):
diff = set(list_1).difference(set(list_2))
occurrences = {i: list_1.count(i) for i in diff}.items()
return [elems[0] for elems in sorted(occurrences, key = lambda pair: pair[1])]
Output:
[1, 3, 2, 4]
CodePudding user response：
If you're OK with the default order for tied counts, this is quite succinct:
from collections import Counter
list_1 = [1,1, 2,2,2, 3,3, 4, 4, 4 ,4, 4, 5,5,5,5]
list_2 = [5, 5, 5, 6, 6, 7]
blacklist = set(list_2)
c = Counter(i for i in list_1 if i not in blacklist)
[n for n,_ in reversed(c.most_common())]
The output is [3, 1, 2, 4] (note the counts for 1 and 3 are the same).
If you can rely on the ordering of your list_1 to be as it is in your example (all clusters are consecutive) you can do a bit better using itertools.groupby:
import itertools.groupby
c = sorted((len(list(g)), i) for i, g in itertools.groupby(list_1) if i not in blacklist)
[i for _, i in c]
Output is [1, 3, 2, 4]
CodePudding user response：
You already have it sorted. So you need to extract the first element?
print([x[0] for x in b])
CodePudding user response：
My interpretation of the question leads me to this:
list_1 = [1,1, 2,2,2, 3,3, 4, 4, 4 ,4, 4, 5,5,5,5]
list_2 = [5, 5, 5, 6, 6, 7]
d = dict()
list_2_s = set(list_2)
for e in list_1:
if e not in list_2_s:
d[e] = d.setdefault(e, 0) 1
lst = [t[1] for t in sorted((v, k) for k, v in d.items())]
print(lst)
Output:
[1, 3, 2, 4]
CodePudding user response：
list(dict.fromkeys(list_1)) this is working same as set but keep order of elements
def sorted_nums(list_1,list_2):
return tuple(filter(lambda x: x not in set(list_2), list(dict.fromkeys(list_1))))